Problem: Is ${428806}$ divisible by $3$ ?
Solution: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {428806}= &&{4}\cdot100000+ \\&&{2}\cdot10000+ \\&&{8}\cdot1000+ \\&&{8}\cdot100+ \\&&{0}\cdot10+ \\&&{6}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {428806}= &&{4}(99999+1)+ \\&&{2}(9999+1)+ \\&&{8}(999+1)+ \\&&{8}(99+1)+ \\&&{0}(9+1)+ \\&&{6} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {428806}= &&\gray{4\cdot99999}+ \\&&\gray{2\cdot9999}+ \\&&\gray{8\cdot999}+ \\&&\gray{8\cdot99}+ \\&&\gray{0\cdot9}+ \\&& {4}+{2}+{8}+{8}+{0}+{6} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${428806}$ is divisible by $3$ if ${ 4}+{2}+{8}+{8}+{0}+{6}$ is divisible by $3$ Add the digits of ${428806}$ $ {4}+{2}+{8}+{8}+{0}+{6} = {28} $ If ${28}$ is divisible by $3$ , then ${428806}$ must also be divisible by $3$ ${28}$ is not divisible by $3$, therefore ${428806}$ must not be divisible by $3$.